Electronic – Will using a wallpowered usb cable to power a beaglebone black provide more power than on USB to computer

My understanding of the your design is that the entire device is on a single PCB, is within a single enclosure, and is connected to the host by a single USB cable. You've integrated a hub onto the PCB to allow both the devices to communicate with the PC. The following answer will hinge on these assumptions, if it's made of several separate devices connected by disconnectable cables then that changes things.

In this case, I suggest that you simply configure the hub to enumerate as a high-power device, and share the resulting 500 mA among the whole board. Interestingly enough, TI's ganged-port sample schematic shows the devices all connected together, even when using their power management IC:

The incoming 5V power supply line (highlighted in blue, as it's one of two nets that we're interested in on this complicated schematic) is connected to a TPS2041 power management IC (a generous description, it's really just a FET that shuts down when it detects 500mA of current being passed). However, each of the inputs are shorted together, and each of the outputs are shorted together as well, and then distributed to each of the downstream ports (the net shown in red).

Basically, they're doing overcurrent protection for all of the downstream sections in a single IC. They have no way of detecting whether they have three low-power (100mA) units, a single high-power unit, or two low-power units and one 300 mA unit. All these options are acceptable based on this reference design. You wrote:

According to the USB specification, a bus-powered hub can provide only one unit per downstream port while drawing max 5 units...

but, to directly answer your question, this design from Texas Instruments (a USB group member and major implementor) shows that you only have to guarantee that the total current is less than 5 units.

To solve your problem, the rules state (taken from the excellent USB in a nutshell document):

High power bus powered functions will draw all its power from the bus and cannot draw more than one unit load until it has been configured, after which it can then drain 5 unit loads (500 mA Max) provided it asked for this in its descriptor.

If you can guarantee that your driver stage will not begin drawing current until the device has been configured (which might be as simple as a timed delay in the host controller), you can simply wire everything together. Because your entire circuit is on a single PCB and has no user-accessible downstream ports, you can probably also leave out the TPS2041 and simply design the system to not require more than 500 mA of current in any state.

Another benefit of enumerating as a high-power device is improved input voltage specifications. When you have enumerated as a low-power device, the host is only required to produce 4.40 V at the upstream port (which will be lower at your device due to the resistance of the cable). When you have enumerated as a high-power device, the specification guarantees that you'll get 4.75 V, which is more likely to be within the operating range of any 5V components you may be using.

To try to get closure on this, and for my own sanity, I read Section 5.4 "Power Management" of the the BeagleBone Black System Reference Manual (BBB_SRM), Rev-C.1.

It says BBB uses two devices to supply the board with power:
TPS65217C has a Vin (Max) 5.8V
LDO TLV70233 has a Vin (Max) 5.5V
These are operating voltages.

The BBB_SRM says the "external LDOTLV70233 provides the 3.3V rail for the rest of the board."

However, searching the BeagleBone Black Schematic (BBB_SCH), Rev C, does not find the TLV70233.

There is a TL5209 on the "BeagleBone Black Power Management" schematic. It is connected to an output of the TPS65217C. It is not connected to the input power socket, and hence isn't directly constraining the maximum DC voltage input. It provides 'VDD_3V3B', so I believe it is the LDO referred to in section 5.4 of the BBB_SRM.

The text of section 5.4 is the same as the Rev-A6 BBB_SRM, so I suspect the text of the BBB_SRM is out of synch with the BBB_SCH Rev-C.

Summary @Farham is correct.
The only active part connected to the DC socket is the TPS65217C, and it has a maximum operating voltage of 5.8V. So 5.24V should be fine.

I assume the BeagleBone Black engineers specify The DC supply should be well regulated and 5V +/-.25V to provide some headroom; the BBB power management subsystem will survive voltages more than 10% over nominal.

Edit: I have raised an issue about the inconsistency of LDO part number of section 5.4 of the BBB_SRM and the "BeagleBone Black Power Management" schematic BBB_SCH at BeagleBone Black github